At a temperature of 60°F, a 0.03-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=α=12.5 x 10-6/°F] bar with a width of 3.0 in. and a thickness of 0.70 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.9 in. and a thickness of 0.70 in. The supports at A and C are rigid. Assume h1=3.0 in., h2=1.9 in., L1=27 in., L2=47 in., and Δ=Δ= 0.03 in. Determine: (1)L₁h₁BAh₂✓(2)L2 Find the force in the Bar (1), F₁, and the force in Bar (2), F2, at a temperature of 260°F. By convention, a tension force is positive and a compression force is negative.Answers:F₁ =F₂=Part 6Answers:Find 0₁ and 02, the normal stresses in Bars (1) and (2), respectively. By convention, a tension stress is positive and a compression stress is negative.0₁ =0₂ =Part 7Answers:kipsd₁ =kipsd2 =Determine 8₁ and 82, the deformations of Bars (1) and (2), respectively. By convention, a deformation is positive for a member that elongates, and it is negative for a member that is shortened.ksiksiin.in.

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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At a temperature of 60°F, a 0.03-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=α=12.5 x 10^-6/°F] bar with a width of 3.0 in. and a thickness of 0.70 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10^-6/°F] bar with a width of 1.9 in. and a thickness of 0.70 in. The supports at A and C are rigid. Assume h₁=3.0 in., h₂=1.9 in., L₁=27 in., L₂=47 in., and Δ=Δ= 0.03 in. Determine:

Find the force in the Bar (1), F₁, and the force in Bar (2), F2, at a temperature of 260°F. By convention, a tension force is positive and a compression force is negative.
Answers:
F₁ =
F₂=
Part 6
Answers:
Find 0₁ and 02, the normal stresses in Bars (1) and (2), respectively. By convention, a tension stress is positive and a compression stress is negative.
0₁ =
0₂ =
Part 7
Answers:
kips
d₁ =
kips
d2 =
Determine 8₁ and 82, the deformations of Bars (1) and (2), respectively. By convention, a deformation is positive for a member that elongates, and it is negative for a member that is shortened.
ksi
ksi
in.
in.

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